Proof by Contradiction | 反证法

📚 Proof by Contradiction | 反证法

Proof by contradiction is a powerful method of mathematical proof in which we assume the opposite of what we want to prove, then show that this assumption leads to an impossible or absurd result. Since mathematics must be consistent, the original statement must therefore be true. In Edexcel A-Level Mathematics, you are expected to use this technique to prove irrationality, the infinitude of primes, and statements about parity or divisibility.

反证法是一种强有力的数学证明方法：我们先假设要证明的结论的反面成立，然后推出一个不可能或荒谬的结果。由于数学体系必须自洽，因此原命题必定为真。在 Edexcel A-Level 数学中，你需要运用反证法证明无理数、素数无穷多以及关于奇偶性或整除性的命题。

1. What is Proof by Contradiction? | 什么是反证法？

Proof by contradiction is a method of establishing the truth of a statement by showing that assuming the opposite leads to an impossible situation. In formal terms, to prove a proposition P, we begin with the assumption ‘not P’. If that assumption logically forces a contradiction, then ‘not P’ cannot be true, so P must be true. This method is also called indirect proof or reductio ad absurdum.

反证法是一种通过证明假设反面会导致不可能情况来确立命题真实性的方法。形式上，要证明命题 P，我们先假设 “非 P”。如果这个假设在逻辑上推出了矛盾，那么 “非 P” 不可能成立，因此 P 必定为真。这种方法也称为间接证明或归谬法。

2. The Logical Structure of Contradiction Proofs | 反证法的逻辑结构

Every proof by contradiction follows the same skeleton: state the proposition, assume its negation, derive consequences of that negation, reach a contradiction, and conclude the original proposition. The contradiction can take many forms, such as 0 = 1, an integer being both even and odd, or a number being both rational and irrational.

每个反证法证明都遵循相同的骨架：陈述命题，假设其否定，从该否定推出结论，达到一个矛盾，然后断定原命题成立。矛盾可以有多种形式，例如 0 = 1、一个整数既是偶数又是奇数，或者一个数既是有理数又是无理数。

Assume ¬P → deduce a contradiction → conclude P

In an Edexcel exam, you should always write down the assumption clearly, using a phrase such as ‘Assume, for contradiction, that …’ so the examiner can follow your logic.

在 Edexcel 考试中，你应当始终清楚地写下假设，使用 “Assume, for contradiction, that …” 这样的表述，以便阅卷老师能跟上你的逻辑。

3. Worked Example 1: Proving √2 Is Irrational | 例题 1：证明 √2 是无理数

Suppose, for contradiction, that √2 is rational. Then √2 = a/b, where a and b are integers with no common factors and b ≠ 0.

假设 √2 是有理数，那么 √2 = a/b，其中 a 和 b 是没有公因数的整数，且 b ≠ 0。

Squaring both sides gives 2 = a²/b², so a² = 2b². This means a² is even. Since the square of an odd integer is odd, a must be even. Let a = 2k for some integer k.

两边平方得到 2 = a²/b²，所以 a² = 2b²。这意味着 a² 是偶数。由于奇数的平方是奇数，所以 a 必为偶数。设 a = 2k，其中 k 为整数。

√2 = a/b → a² = 2b² → a = 2k → b² = 2k² → contradiction

Substituting a = 2k into a² = 2b² gives (2k)² = 2b², so 4k² = 2b², hence b² = 2k². Therefore b² is even, and by the same argument b must be even.

将 a = 2k 代入 a² = 2b² 得到 (2k)² = 2b²，所以 4k² = 2b²，因此 b² = 2k²。于是 b² 是偶数，同理 b 必为偶数。

Now both a and b are even, so they share a common factor of 2. This contradicts the assumption that a/b is in its lowest terms. Hence √2 cannot be rational.

现在 a 和 b 都是偶数，因此它们有公因数 2。这与假设 a/b 是最简分数矛盾。因此 √2 不可能是有理数。

4. Worked Example 2: Proving There Are Infinitely Many Primes | 例题 2：证明素数有无穷多个

Assume, for contradiction, that there are finitely many primes: p₁, p₂, …, pₙ. Consider the number N = p₁ × p₂ × … × pₙ + 1.

假设素数只有有限个：p₁, p₂, …, pₙ。考虑数 N = p₁ × p₂ × … × pₙ + 1。

Since N is greater than every listed prime, N is either prime or composite. If N is prime, then it is a prime that is not in the original list, which contradicts the assumption that the list contains all primes.

由于 N 大于列表中的每一个素数，N 要么是素数，要么是合数。如果 N 是素数，那么它是一个不在原列表中的素数，这与列表包含所有素数的假设矛盾。

If N is composite, it must have a prime factor q. But dividing N by any listed prime pᵢ leaves remainder 1, so q cannot be any of p₁, p₂, …, pₙ. This again contradicts the assumption that all primes are in the list.

如果 N 是合数，它必定有一个素因数 q。但 N 除以任意列表中的素数 pᵢ 都余 1，因此 q 不可能是 p₁, p₂, …, pₙ 中的任何一个。这再次与列表包含所有素数的假设矛盾。

N = p₁ × p₂ × … × pₙ + 1 → remainder 1 for every pᵢ → contradiction

Therefore the assumption that there are finitely many primes is false, so there must be infinitely many primes.

因此，素数有限的假设是错误的，所以素数必定有无穷多个。

5. Worked Example 3: Parity Proofs – If n² Is Even, Then n Is Even | 例题 3：奇偶性证明——若 n² 为偶数，则 n 为偶数

We want to prove that for any integer n, if n² is even, then n is even. Suppose, for contradiction, that n² is even but n is odd.

我们要证明：对任意整数 n，若 n² 为偶数，则 n 为偶数。假设 n² 是偶数，但 n 是奇数。

If n is odd, then n = 2k + 1 for some integer k. Squaring gives n² = (2k + 1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1, which is odd.

如果 n 是奇数，那么 n = 2k + 1，其中 k 为整数。平方得到 n² = (2k + 1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1，这是一个奇数。

n² = (2k + 1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 → odd

This contradicts the given fact that n² is even. Therefore n cannot be odd, so n must be even.

这与已知 n² 为偶数矛盾。因此 n 不可能是奇数，所以 n 必为偶数。

6. Worked Example 4: Proving No Largest Integer Exists | 例题 4：证明不存在最大整数

Prove that there is no largest positive integer. Suppose, for contradiction, that there is a largest positive integer N. Since N is a positive integer, N + 1 is also a positive integer.

证明不存在最大正整数。假设存在一个最大正整数 N。由于 N 是正整数，N + 1 也是正整数。

N + 1 > N

But N + 1 > N, which contradicts the assumption that N is the largest positive integer. Therefore no largest positive integer exists.

但是 N + 1 > N，这与 N 是最大正整数的假设矛盾。因此不存在最大正整数。

7. Worked Example 5: Proving No Smallest Positive Rational Exists | 例题 5：证明不存在最小正有理数

Prove that there is no smallest positive rational number. Suppose, for contradiction, that r is the smallest positive rational number.

证明不存在最小正有理数。假设 r 是最小正有理数。

Then r/2 is also a positive rational number, and r/2 < r. This contradicts the assumption that r is the smallest positive rational number.

那么 r/2 也是正有理数，并且 r/2 < r。这与 r 是最小正有理数的假设矛盾。

r/2 < r → contradiction

Therefore no smallest positive rational number exists.

因此不存在最小正有理数。

8. Common Mistakes in Proof by Contradiction | 反证法常见错误

A common mistake is failing to state the negation of the proposition clearly. If the assumption is ambiguous, the proof becomes difficult to follow and may not be accepted by examiners.

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Proof by Contradiction | 反证法